Equliateral Triangle inscribed in circle

**hachiko** · February 6th 2010, 11:59 AM

Equilateral triangle ABC is inscribed in a circle. The angle bisector at A intersects the circle again at P (AP is the diameter). If the length of segment AC is 10, find AP - (BP + CP).

I will upload a picture soon, but help is greatly appreciated!!

**Opalg** · February 6th 2010, 12:43 PM

Originally Posted by hachiko

Equilateral triangle ABC is inscribed in a circle. The angle bisector at A intersects the circle again at P (AP is the diameter). If the length of segment AC is 10, find AP - (BP + CP).

Let O be the centre of the circle. Can you see why OP = BP = CP? What does that tell you about AP – (BP + CP)?

**Soroban** · February 6th 2010, 03:27 PM

Hello, hachiko!

Another approach . . .

Equilateral triangle $ABC$ is inscribed in a circle.
The angle bisector at $A$ intersects the circle again at $P$ ( $AP$ is adiameter).
If the length of segment $AC$ is 10, find: . $AP - (BP + CP)$

Code:

                A
              * o *
          *    /|\    *
        *     / | \     *
       *     /  |  \ 10  *
            /   |30°\
      *    /    |    \    *
      *   /     |     \   *
      *  /      |      \  *
        /       |       \
     B o- - - - + - - - -o C
        *       |     * *
          * 60° |  *  *
              * o *
                P

We have: inscribed equilateral triangle $ABC\!:\;\;AC = 10$
. . with diameter $AP,\;\;\angle CAP = 30^o.$
Draw chord $CP$ .

$\Delta ACP$ is inscribed in a semicircle.
. . Hence: . $\angle ACP = 90^o,\;\;\angle APC = 60^o$

We have: . $CP = x,\;\;AC = 10,\;\;AP = 2x$ .(30°-60° right triangle)

Pythagorus says: . $CP^2 + AC^2 \:=\:AP^2 \quad\Rightarrow\quad x^2 + 10^2 \:=\:(2x)^2$

. . $x^2 + 100 \:=\:4x^2 \quad\Rightarrow\quad 3x^2 \:=\:100 \quad\Rightarrow\quad x^2 \:=\:\frac{100}{3}\quad\Rightarrow\quad x \:=\:\frac{10\sqrt{3}}{3}$

Hence: . $BP \,=\,CP \,=\,\frac{10\sqrt{3}}{3},\;\;AP \,=\,\frac{20\sqrt{3}}{3}$

Therefore: . $AP - (BP + CP) \;\;=\;\;\frac{20\sqrt{3}}{3} - \left(\frac{10\sqrt{3}}{3} + \frac{10\sqrt{3}}{3}\right) \;\;=\;\;0$

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